Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/ 4 8
/ / 11 13 4
/ \ 7 2 1
要求求出是否有一条路径和与给出的值相等，注意中间节点与叶子节点的判断：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool hasPathSum(TreeNode* root, int sum) { 13 if(root == NULL) return false; 14 return checkSum(root, sum); 15 } 16 17 bool checkSum(TreeNode* root, int sum) 18 { 19 if(root != NULL && sum == root->val && root->left == NULL && root->right == NULL){ 20 return true;//上面这个判断确实是叶子节点，值也同时满足 21 } 22 else if(root == NULL) 23 return false; 24 else 25 return checkSum(root->left, sum - root->val) || checkSum(root->right, sum - root->val); 26 } 27 };
原文：http://www.cnblogs.com/-wang-cheng/p/4904786.html